What is the integral from 1 to 3 of 1/(x^2+6x+10) ?

The integral from 1 to 3 of 1/(x^2+6x+10) is arctan(6)-arctan(4)

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integral from 1 to 3 of 1/(x^2+6x+10)

\int_{1}^{3} \frac{1}{x ^{2} + 6 x + 10} d x

arctan (6) - arctan (4)

Solution steps

\int_{1}^{3} \frac{1}{x ^{2} + 6 x + 10} d x

Complete the square

x^{2} + 6 x + 10 : (x + 3)^{2} + 1

= \int_{1}^{3} \frac{1}{( x + 3 ) ^{2} + 1} d x

Apply u-substitution

: \int_{4}^{6} \frac{1}{u ^{2} + 1} d u

= \int_{4}^{6} \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= [arctan (u)]_{4}^{6}

Compute the boundaries:

arctan (6) - arctan (4)

= arctan (6) - arctan (4)

\int_{- 1}^{0} x sin (nπ x) d x

\int_{2}^{3} \frac{2}{3 x - 1} d x

\int_{- 5}^{2} x^{2} + 3 x - 10 d x

\int_{1}^{4} 5 x^{2} + x d x

\int_{0}^{4} e^{- 2 x} d x

What is the integral from 1 to 3 of 1/(x^2+6x+10) ?
The integral from 1 to 3 of 1/(x^2+6x+10) is arctan(6)-arctan(4)